A290616 -id:A290616 - cp's OEIS frontend

A292320 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - S^3.

1, 1, 2, 4, 6, 12, 22, 36, 67, 122, 209, 377, 681, 1193, 2130, 3823, 6764, 12043, 21531, 38252, 68076, 121456, 216126, 384691, 685636, 1220767, 2173346, 3871747, 6894873, 12276852, 21866387, 38941846, 69344928, 123500513, 219943018, 391676701, 697538335

Offset: 0

Clark Kimberling, Sep 14 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A291728 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1, 0, 4, -2, 0, -3, 1, 0, 1)

Cf. A079978, A290616.

z = 60; s = x/(x - x^3); p = 1 - s - s^3;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A292320 *)

G.f.: -((1 + x^2 - 2 x^3 + x^6)/(-1 + x + 4 x^3 - 2 x^4 - 3 x^6 + x^7 + x^9)).

a(n) = a(n-1) + 4*a(n-3) - 2*a(n-4) - 3*a(n-6) + a(n-7) + a(n-9) for n >= 10.

A291035 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - 2 S^2.

Original entry on oeis.org

1, 3, 5, 12, 27, 58, 130, 285, 629, 1389, 3060, 6753, 14892, 32844, 72445, 159775, 352401, 777244, 1714259, 3780930, 8339090, 18392449, 40565829, 89470733, 197333940, 435233685, 959938112, 2117210180, 4669654005, 10299246171, 22715702489, 50101058976

Offset: 0

Clark Kimberling, Sep 14 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A291728 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,2,-1,0,-1).

Cf. A079978, A289918, A290616.

z = 60; s = x/(x - x^3); p = 1 - s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978 *)
Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291035 *)

G.f.: -(1 + x)*(-1 - x + x^2)/((-1 - x + x^3)*(-1 + 2*x + x^3)).

a(n) = a(n-1) + 2*a(n-2) + 2*a(n-3) - a(n-4) - a(n-6) for n >= 7.

A291036 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - 2 S - 2 S^2.

Original entry on oeis.org

2, 6, 16, 46, 132, 376, 1074, 3066, 8752, 24986, 71328, 203624, 581298, 1659462, 4737360, 13524006, 38607732, 110215648, 314638754, 898216794, 2564189568, 7320134930, 20897197344, 59656394448, 170304435554, 486177568038, 1387918211824, 3962167507006

Offset: 0

Clark Kimberling, Sep 14 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A291728 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2, 2, 2, -2, 0, -1)

Cf. A079978, A290616, A291037.

z = 60; s = x/(x - x^3); p = 1 - 2 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291036 *)
u/2 (* A291037 *)

G.f.: -((2 (-1 - x + x^3))/(1 - 2 x - 2 x^2 - 2 x^3 + 2 x^4 + x^6)).

a(n) = 2*a(n-1) + 2*a(n-2) + 2*a(n-3) - 2*a(n-4) - a(n-6) for n >= 7.

A291037 a(n) = (1/2)*A291036(n).

Original entry on oeis.org

1, 3, 8, 23, 66, 188, 537, 1533, 4376, 12493, 35664, 101812, 290649, 829731, 2368680, 6762003, 19303866, 55107824, 157319377, 449108397, 1282094784, 3660067465, 10448598672, 29828197224, 85152217777, 243088784019, 693959105912, 1981083753503, 5655510252642

Offset: 0

Clark Kimberling, Sep 14 2017

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (2, 2, 2, -2, 0, -1)

Cf. A079978, A290616, A291036.

z = 60; s = x/(x - x^3); p = 1 - 2 s - 2 s^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291036 *)
u/2 (* A291037 *)
LinearRecurrence[{2,2,2,-2,0,-1},{1,3,8,23,66,188},30] (* Harvey P. Dale, Sep 24 2017 *)

G.f.: -((-1 - x + x^3)/(1 - 2 x - 2 x^2 - 2 x^3 + 2 x^4 + x^6)).

a(n) = 2*a(n-1) + 2*a(n-2) + 2*a(n-3) - 2*a(n-4) - a(n-6) for n >= 7.

A291038 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - 2 S)^2.

Original entry on oeis.org

4, 12, 32, 84, 216, 544, 1348, 3300, 8000, 19236, 45936, 109056, 257604, 605820, 1419232, 3313396, 7711944, 17900320, 41445764, 95746260, 220735616, 507934276, 1166792160, 2676017408, 6128381316, 14015556588, 32012831648, 73033858964, 166434905016

Offset: 0

Clark Kimberling, Sep 14 2017

Suppose s = (c(0), c(1), c(2), ...) is a sequence and p(S) is a polynomial. Let S(x) = c(0)*x + c(1)*x^2 + c(2)*x^3 + ... and T(x) = (-p(0) + 1/p(S(x)))/x. The p-INVERT of s is the sequence t(s) of coefficients in the Maclaurin series for T(x). Taking p(S) = 1 - S gives the "INVERT" transform of s, so that p-INVERT is a generalization of the "INVERT" transform (e.g., A033453).

See A291728 for a guide to related sequences.

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, -4, 2, -4, 0, -1)

Cf. A079978, A290616, A291039.

z = 60; s = x/(x - x^3); p = (1 - 2 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291038 *)
u/4 (* A291039 *)

G.f.: -((4 (-1 + x + x^3))/(-1 + 2 x + x^3)^2).

a(n) = 4*a(n-1) - 4*a(n-2) + 2*a(n-3) - 4*a(n-4) - a(n-6) for n >= 7.

A291039 a(n) = (1/4)*A291038(n).

Original entry on oeis.org

1, 3, 8, 21, 54, 136, 337, 825, 2000, 4809, 11484, 27264, 64401, 151455, 354808, 828349, 1927986, 4475080, 10361441, 23936565, 55183904, 126983569, 291698040, 669004352, 1532095329, 3503889147, 8003207912, 18258464741, 41608726254, 94722900936, 215428701233

Offset: 0

Clark Kimberling, Sep 14 2017

Clark Kimberling, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (4, -4, 2, -4, 0, -1)

Cf. A079978, A290616, A291038.

z = 60; s = x/(x - x^3); p = (1 - 2 s)^2;
Drop[CoefficientList[Series[s, {x, 0, z}], x], 1]  (* A079978  *)
u = Drop[CoefficientList[Series[1/p, {x, 0, z}], x], 1]  (* A291038 *)
u/4 (* A291039 *)

G.f.: -((-1 + x + x^3)/(-1 + 2 x + x^3)^2).

a(n) = 4*a(n-1) - 4*a(n-2) + 2*a(n-3) - 4*a(n-4) - a(n-6) for n >= 7.

cp's OEIS Frontend

A292320 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - S^3.

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A291035 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - S - 2 S^2.

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A291036 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = 1 - 2 S - 2 S^2.

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A291037 a(n) = (1/2)*A291036(n).

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A291038 p-INVERT of (1,0,0,1,0,0,1,0,0,...), where p(S) = (1 - 2 S)^2.

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A291039 a(n) = (1/4)*A291038(n).

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